Write . A double covering has a deck transformation that exchanges the two points in every fibre. Every singular simplex has exactly two lifts and . Define the mod-two transfer chain map of a double covering byand let send a simplex of to its composite with . Uniqueness of lifted faces shows that commutes with the boundary operator, so both maps are chain maps.
For each base simplex, its two lifts span a copy of . On this summand, is and is . Hence the sequence is exact in every degree:
Assume for contradiction that the involution is fixed-point-free. The finite group action is then a covering space action, so the quotient mapis a double covering and is an -manifold.
Apply the long exact sequence in homology to the transfer chain map of a double covering. Since is contractible, its positive-dimensional mod-two homology vanishes. The degree-zero portion, together with the fact that is an isomorphism, givesIn every higher degree the same exact sequence givesThus for every . This contradicts homology above the dimension of a manifold, which gives for . Therefore has a fixed point.
A closed manifold of odd dimension has Euler characteristic zero by the Euler characteristic of an odd-dimensional closed manifold. Decomposing into and the removed closed ball along givesso .
If the antipodal boundary map extended to a fixed-point-free involution of , the quotient map would be a double covering. The formula for Euler characteristic under a finite covering would implywhich is impossible. Hence no such extension exists.
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